Some Interesting Properties of 39

My daughter-in-law turned 39 yesterday and so I was prompted to investigate some of its mathematical properties. One of its properties is its membership in OEIS A055233:

A055233: composite numbers equal to the sum of the primes from their smallest prime factor to their largest prime factor.

The only members of this sequence in the range up to 40000 are 10, 39, 155 and 371. All are semiprimes and factorise as follows:

• \(10 = 2 \times 5 \text{ with } 2 + 3 + 5 = 10 \)
• \(39 = 3 \times 13 \text{ with } 3 + 5 + 7 + 11+13 = 39\)
• \(155 = 5 \times 31 \text{ with } 5 + 7 + \ldots + 29 + 31=155\)
• \(371 = 7 \times 53 \text{ with } 7 + 11 + \ldots + 47 + 53=371\)

Because they are semiprimes they are thus equal to the product of their smallest and largest prime factors. However, this is not the case for the next member of the sequence: 2935561623745. The reason is that it is not a semiprime.

• \(2935561623745= 5 \times 19 \times 53 \times 61 \times 9557887\)

The next member of the sequence 454539357304421 is a semiprime and thus follows the pattern of the first four members of the sequence:

• \(454539357304421 = 3536123 \times 128541727\)

So we see that 39 by virtue of its membership in OEIS A055233 is rather special. Of course, it has some other interesting qualities. For example, it can be constructed from the first three powers of 3:$$39=3+3^2+3^3$$Gemini also mentions the following number properties:

Beyond these patterns, 39 is also classified as a \( \textbf{Perrin number}\) and a \( \textbf{Størmer number}\), placing it within specialized mathematical sequences that are far from intuitive. 

The number also has an \( \textbf{aliquot sum}\) of 17, which is a prime number, a unique characteristic that links it to a specific aliquot sequence. 

In the realm of number partitions, 39 is notable as the smallest natural number to have three distinct partitions into three parts that all yield the same product, 1200. These partitions are:

• {25, 8, 6} 

• {24, 10, 5} 

• {20, 15, 4}. 

Lastly, in analytic number theory, the \( \textbf{Mertens function}\) returns a value of 0 when given 39, a property that suggests a form of numerical equilibrium or stability, a concept that finds intriguing parallels in other domains. See blog post Zeroes of the Mertens Function.

39 is also what's termed a \( \textbf{perfect totient number} \) because the sum of its iterated totients equals the number itself. Let's confirm this:$$ \begin{align} \phi(39) &=24 \\ \phi(24) &=8 \\ \phi(8) &=4 \\ \phi(4) &=2 \\ \phi(2) &=1 \end{align} $$The sum of these iterated totients equals 39:$$24 + 8 + 4 + 2 + 1 =39$$The perfect totient numbers are listed in OEIS A082897 (permalink):

3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571, 6561, 8751, 15723, 19683, 36759, 46791, 59049, 65535, 140103, 177147, 208191, 441027, 531441, 1594323, 4190263, 4782969, 9056583, 14348907, 43046721

Tricky Entrance Exam Questions

I came across this problem on a YouTube channel. The problem was purported to be a Harvard University entrance exam question.$$ \text{Simplify } \sqrt{\sqrt{121}-\sqrt{120}}$$Once you see the method, it's easy enough so let's start to simplify:$$

\begin{align}

\sqrt{\sqrt{121}-\sqrt{120}} &= \sqrt{11 - 2 \cdot \sqrt{30}} \\

&= \sqrt{11 - 2 \cdot \sqrt{6} \cdot \sqrt{5}} \\

&= \sqrt{6 - 2 \cdot \sqrt{6} \cdot \sqrt{5} + 5} \\

&= \sqrt{(\sqrt{6})^2 - 2 \cdot \sqrt{6} \cdot \sqrt{5} + (\sqrt{5})^2} \\

&= \sqrt{(\sqrt{6} - \sqrt{5})^2} \\

&= \sqrt{6} - \sqrt{5}

\end{align}

$$Here's another one:$$ \text{Simplify } \sqrt{\sqrt{36}-\sqrt{20}}$$The approach is exactly the same:

$$
\begin{align}

\sqrt{\sqrt{36}-\sqrt{20}} &= \sqrt{6 - 2 \cdot \sqrt{5}} \\

&= \sqrt{6- 2 \cdot \sqrt{5} \cdot \sqrt{1}} \\

&= \sqrt{5 - 2 \cdot \sqrt{5} \cdot \sqrt{1} + 1} \\

&= \sqrt{(\sqrt{5})^2 - 2 \cdot \sqrt{5} \cdot \sqrt{1} + (\sqrt{1})^2} \\

&= \sqrt{(\sqrt{5} - \sqrt{1})^2} \\

&= \sqrt{5} - \sqrt{1}\\
&=\sqrt{5}-1

\end{align}

$$

A Special Date

 I came across this article today that discusses today's date: the 16th of September 2025:

Once a century, a very special day comes along. That day is today — 9/16/25.

Pi Day (3/14) often comes with sweet treats; Square Root Day (4/4/16 or 5/5/25, for example) has a certain numerical rhyme. But the particular string of numbers in today's date may be especially delightful to the brains of mathematicians and the casual nerds among us.

First, "all three of the entries in that date are perfect squares — and what I mean by that is \(9\) is equal to \(3^2\), \(16\) is equal to \(4^2\), and \(25\) is equal to \(5^2,\)" says Colin Adams, a mathematician at Williams College who was first tipped off about today's special qualities during a meeting with his former student, Jake Malarkey.

Next, those perfect squares come from consecutive numbers — three, four, and five.

But perhaps most special of all is that three, four, and five are an example of what's called a Pythagorean triple.

"And what that means," explains Adams, "is that if I take the sum of the squares of the first two numbers, \(3^2 + 4^2\), which is \(9 + 16\) is equal to \(25\), which is \(5^2\), so \(3^2 + 4^2 = 5^2\)."

This is the Pythagorean Theorem: \(a^2 + b^2 = c^2\). "And that in fact is the most famous theorem in all of mathematics," says Adams.

It's a theorem that means something geometrically, too. Any Pythagorean triple — including 3, 4, and 5 — also gives the lengths of the three sides of a right triangle. That is, the squares of the two shorter lengths add up to the square of the final, longer side (the hypotenuse).

There are no other dates this century that meet all these conditions, so most of us will experience it just once in our lifetime.

(Fun bonus: It turns out the full year, \(2025\), is also a perfect square: \(45 \times 45\).)

In any case, Adams says that if it were up to him, he'd call the day Pythagorean Triple Square Day. And he plans on celebrating with a rectangular cake cut along the diagonal to yield two right triangles.

"If I have any luck at all, if I can find a cake with the right dimensions, it'll look like a 3, 4, 5 cake, namely edge length 3, edge length 4, and edge length 5," he says. In the middle, he intends to have the date inscribed in icing.

"This date is hiding one of the most beautiful coincidences we will ever encounter," says Terrence Blackman, chair of the mathematics department at Medgar Evers College in the City University of New York. "Those numbers, they tell a story that goes back to ancient Greece."

Blackman says the Pythagorean Theorem is used frequently by carpenters and architects. But for him, as a mathematician, today's date captures a special elegance.

"It reveals some kind of hidden mathematical poetry that is sitting there — just like walking and coming upon a beautiful flower," he says.

In a world that can feel chaotic, Blackman feels that a day like today shows that math can provide a source of comfort.

"It reminds us that beauty and meaning can be found anywhere and everywhere," he says. "We just have to continue to look for it."

Sequences Involving SOD and POD

The number associated with my diurnal age today, 27923, has the interesting property that its sum of digits (23) is equal to the last two digits of the number. This number is part of a sequence of consecutive numbers that all share this same property. The numbers are 27920 up to 27929. In the range of numbers up to 40000, there are 440 numbers with this property. They are (permalink):

SOD = Concatenation of Last Two Digits

910, 911, 912, 913, 914, 915, 916, 917, 918, 919, 1810, 1811, 1812, 1813, 1814, 1815, 1816, 1817, 1818, 1819, 2710, 2711, 2712, 2713, 2714, 2715, 2716, 2717, 2718, 2719, 3610, 3611, 3612, 3613, 3614, 3615, 3616, 3617, 3618, 3619, 4510, 4511, 4512, 4513, 4514, 4515, 4516, 4517, 4518, 4519, 5410, 5411, 5412, 5413, 5414, 5415, 5416, 5417, 5418, 5419, 6310, 6311, 6312, 6313, 6314, 6315, 6316, 6317, 6318, 6319, 7210, 7211, 7212, 7213, 7214, 7215, 7216, 7217, 7218, 7219, 8110, 8111, 8112, 8113, 8114, 8115, 8116, 8117, 8118, 8119, 9010, 9011, 9012, 9013, 9014, 9015, 9016, 9017, 9018, 9019, 9920, 9921, 9922, 9923, 9924, 9925, 9926, 9927, 9928, 9929, 10810, 10811, 10812, 10813, 10814, 10815, 10816, 10817, 10818, 10819, 11710, 11711, 11712, 11713, 11714, 11715, 11716, 11717, 11718, 11719, 12610, 12611, 12612, 12613, 12614, 12615, 12616, 12617, 12618, 12619, 13510, 13511, 13512, 13513, 13514, 13515, 13516, 13517, 13518, 13519, 14410, 14411, 14412, 14413, 14414, 14415, 14416, 14417, 14418, 14419, 15310, 15311, 15312, 15313, 15314, 15315, 15316, 15317, 15318, 15319, 16210, 16211, 16212, 16213, 16214, 16215, 16216, 16217, 16218, 16219, 17110, 17111, 17112, 17113, 17114, 17115, 17116, 17117, 17118, 17119, 18010, 18011, 18012, 18013, 18014, 18015, 18016, 18017, 18018, 18019, 18920, 18921, 18922, 18923, 18924, 18925, 18926, 18927, 18928, 18929, 19820, 19821, 19822, 19823, 19824, 19825, 19826, 19827, 19828, 19829, 20710, 20711, 20712, 20713, 20714, 20715, 20716, 20717, 20718, 20719, 21610, 21611, 21612, 21613, 21614, 21615, 21616, 21617, 21618, 21619, 22510, 22511, 22512, 22513, 22514, 22515, 22516, 22517, 22518, 22519, 23410, 23411, 23412, 23413, 23414, 23415, 23416, 23417, 23418, 23419, 24310, 24311, 24312, 24313, 24314, 24315, 24316, 24317, 24318, 24319, 25210, 25211, 25212, 25213, 25214, 25215, 25216, 25217, 25218, 25219, 26110, 26111, 26112, 26113, 26114, 26115, 26116, 26117, 26118, 26119, 27010, 27011, 27012, 27013, 27014, 27015, 27016, 27017, 27018, 27019, 27920, 27921, 27922, 27923, 27924, 27925, 27926, 27927, 27928, 27929, 28820, 28821, 28822, 28823, 28824, 28825, 28826, 28827, 28828, 28829, 29720, 29721, 29722, 29723, 29724, 29725, 29726, 29727, 29728, 29729, 30610, 30611, 30612, 30613, 30614, 30615, 30616, 30617, 30618, 30619, 31510, 31511, 31512, 31513, 31514, 31515, 31516, 31517, 31518, 31519, 32410, 32411, 32412, 32413, 32414, 32415, 32416, 32417, 32418, 32419, 33310, 33311, 33312, 33313, 33314, 33315, 33316, 33317, 33318, 33319, 34210, 34211, 34212, 34213, 34214, 34215, 34216, 34217, 34218, 34219, 35110, 35111, 35112, 35113, 35114, 35115, 35116, 35117, 35118, 35119, 36010, 36011, 36012, 36013, 36014, 36015, 36016, 36017, 36018, 36019, 36920, 36921, 36922, 36923, 36924, 36925, 36926, 36927, 36928, 36929, 37820, 37821, 37822, 37823, 37824, 37825, 37826, 37827, 37828, 37829, 38720, 38721, 38722, 38723, 38724, 38725, 38726, 38727, 38728, 38729, 39620, 39621, 39622, 39623, 39624, 39625, 39626, 39627, 39628, 39629

A plot of these numbers shows a quite regular pattern. See Figure 1.

Figure 1: permalink

If we consider the product of digits instead and exclude numbers with the digit 0, then only 31 numbers satisfy in the range up to 40000. These are (permalink):

POD = Concatenation of Last Two Digits

236, 315, 324, 612, 1236, 1315, 1324, 1612, 2136, 2312, 3115, 3124, 3212, 6112, 11236, 11315, 11324, 11612, 12136, 12312, 13115, 13124, 13212, 16112, 21136, 21312, 23112, 31115, 31124, 31212, 32112

Figure 2 shows a graph of these numbers.

Figure 2: permalink

Even Numbers as Sums of Three Pronic Numbers

I saw this problem mentioned in a post on math.stackexchange.com:

See if you can prove that every even number is the sum of three pronic numbers (if needed, we say 0 is also pronic) 

I put the problem to Gemini and got the response that "it's impossible to prove that every even number is the sum of three pronic numbers". I wasn't convinced so I consulted Grok. Here was its response.

Every even non-negative integer can indeed be expressed as the sum of three pronic numbers, where pronic numbers are of the form \(n(n+1)\) for non-negative integers \(n \geq 0\) (thus including 0 as the pronic number for \(n=0\)).

To prove this, first recall that a pronic number \(p_n = n(n+1) = 2 \cdot T_n\), where \(T_n = \frac{n(n+1)}{2}\) is the \(n\)th triangular number (with \(T_0 = 0\)). Let \(E\) be an arbitrary even non-negative integer, so \(E = 2m\) for some non-negative integer \(m\). The claim is equivalent to showing that every non-negative integer \(m\) can be written as the sum of three triangular numbers: \(m = T_a + T_b + T_c\) for some non-negative integers \(a, b, c\). This is a known theorem (attributed to Gauss): every non-negative integer is the sum of at most three triangular numbers. The proof relies on a connection to the three squares theorem and modular arithmetic, as follows. Step 1: Relate triangular numbers to squares.

Observe that \(8T_n + 1 = (2n + 1)^2\). Thus, every number of the form \(8T_n + 1\) is an odd perfect square, and conversely, every odd perfect square is of this form for some \(n\). Step 2: Reformulate the problem.

Suppose \(m = T_a + T_b + T_c\). Multiplying by 8 and adding 3 gives:$$ \begin{align} 8m + 3 &= (8T_a + 1) + (8T_b + 1) + (8T_c + 1) \\ &= (2a + 1)^2 + (2b + 1)^2 + (2c + 1)^2 \end{align} $$so \(8m + 3\) is the sum of three odd squares. The converse also holds: if \(8m + 3\) is the sum of three odd squares, then \(m\) is the sum of three triangular numbers.

Thus, it suffices to show that every number of the form \(8m + 3\) (for \(m \geq 0\)) is the sum of three odd squares. Step 3: Use the three squares theorem

Legendre's three squares theorem states that a positive integer can be expressed as the sum of three squares if and only if it is not of the form \(4^k(8l + 7)\) for non-negative integers \(k, l\). Numbers of the form \(8m + 3\) are congruent to 3 modulo 8. Check if such numbers can be of the forbidden form \(4^k(8l + 7)\):

• For \(k = 0\): \(8l + 7 \equiv 7 \pmod{8}\), not 3.
• For \(k = 1\): \(4(8l + 7) = 32l + 28 \equiv 4 \pmod{8}\), not 3.
• For \(k \geq 2\): \(4^k(8l + 7) \equiv 0 \pmod{8}\), not 3.

Since \(8m + 3\) is never of the forbidden form, every such number is the sum of three squares.

Step 4: Ensure the squares are odd

It remains to show these three squares must be odd. Consider sums of three squares modulo 8. Perfect squares modulo 8 are 0, 1, or 4:

• Odd squares: \(\equiv 1 \pmod{8} \)
• Even squares: \( \equiv 0 or 4 \pmod{8} \)

The possible sums modulo 8 are 0, 1, 2, 3, 4, 5, or 6 (never 7). Specifically for \(\equiv 3 \pmod{8} \)

• Three odd squares: \(1 + 1 + 1 = 3\)
• Two odd, one even: \(1 + 1 + 0 = 2\) or \(1 + 1 + 4 = 6\).
• One odd, two even: \(1 + 0 + 0 = 1, 1 + 0 + 4 = 5\), or \(1 + 4 + 4 = 9 \equiv 1\).
• Three even: \(0 + 0 + 0 = 0, 0 + 0 + 4 = 4\), \( 0 + 4 + 4 = 8 \equiv 0\), or \(4 + 4 + 4 = 12 \equiv 4\).

The only combination yielding \(3 \pmod{8} \) is three odd squares. Therefore, since every \(8m + 3\) is the sum of three squares and \(\equiv 3 \pmod{8}\), it must be the sum of three odd squares. This completes the proof. For any even \(E = 2m\), express \(m = T_a + T_b + T_c\), then$$ \begin{align} E &= 2T_a + 2T_b + 2T_c \\ &= p_a + p_b + p_c \end{align} $$where each \(p_i \) is pronic.

This is a nice straightforward proof. I've written about pronic numbers before in posts titled:

• Even Numbers as Sums and Differences of Two Pronic Numbers 
• Revisiting Sums and Differences of Two Pronic Numbers
• My Yearly Pronic Number
• Pronic Pandigital Numbers and Beyond
• Midway Between Cubics and Pronics
• Some Special Pronic Numbers
• Infinite Sums of Reciprocals of Pronic Numbers

A Range Limit of 40000: Why?

In many of my posts when I'm considering sequences, I only look at sequence members whose values do not exceed 40000. Why? The answer to this question relates to the reason that I set this blog up in the first place. It was designed primarily to post about interesting sequences associated with the numbers marking my diurnal age.

If we divide 40000 by 365.2425 (the average number of days in a year) we get slightly more than 109.5 years and there are not many people who live to that ripe old age. Here are milestones, one might say, on the road to oblivion.

Quite a few people won't see 30000 days let alone 40000 but many will and hence the realistic upper limit to the numbers that I normally investigate. The focus of my blog posts is generally the number associated with my diurnal age and the sequences that it can be connected to. For example, today I am \( \textbf{27920} \) days old.

This number found its way into a sequence that I created that involves gapful numbers with the property that not only does the number formed by the concatenation of the first and last digit divide the number but this concatenated number is also the sum of the number's digits. Thus we have:$$ \begin{align} \frac{27920}{20} &=1396 \\ \\ 2 + 7 + 9 + 2 + 0 &= 20 \end{align}$$I described this sequence in a post titled Gapgul Numbers in December of 2024. Interestingly, 27920 also has the property that it has 20 divisors. Only 19 numbers satisfy this additional criterion in the range up to 40000. The conditions to be met are:

• the number is gapful meaning the number formed by concantenating the first and last digits divides the number without remainder
• the sum of the number's digits equals the number formed by the concatenated first and last digits
• the number of divisors of the number equals its sum of digits (and the concatenated number)

Here are the numbers:

1548, 1812, 1908, 10188, 10548, 11268, 12252, 12612, 12708, 13428, 14052, 14412, 15138, 18108, 21984, 26480, 27920, 29360, 39996

Here are the details:

  number        factor          concat   dividend   S0D   divisors

  1548     2^2 * 3^2 * 43         18       86         18    18
  1812     2^2 * 3 * 151          12       151        12    12
  1908     2^2 * 3^2 * 53         18       106        18    18
  10188    2^2 * 3^2 * 283        18       566        18    18
  10548    2^2 * 3^2 * 293        18       586        18    18
  11268    2^2 * 3^2 * 313        18       626        18    18
  12252    2^2 * 3 * 1021         12       1021       12    12
  12612    2^2 * 3 * 1051         12       1051       12    12
  12708    2^2 * 3^2 * 353        18       706        18    18
  13428    2^2 * 3^2 * 373        18       746        18    18
  14052    2^2 * 3 * 1171         12       1171       12    12
  14412    2^2 * 3 * 1201         12       1201       12    12
  15138    2 * 3^2 * 29^2         18       841        18    18
  18108    2^2 * 3^2 * 503        18       1006       18    18
  21984    2^5 * 3 * 229          24       916        24    24
  26480    2^4 * 5 * 331          20       1324       20    20
  27920    2^4 * 5 * 349          20       1396       20    20
  29360    2^4 * 5 * 367          20       1468       20    20
  39996    2^2 * 3^2 * 11 * 101   36       1111       36    36

6-P-6 Primes And Beyond

What I mean by a 6-P-6 prime is a prime, greater than 2,  whose two adjacent composite numbers contain exactly six prime factors with multiplicity. There are only 30 such primes in the range up to 40000 and they are (permalink):

1889, 3079, 4591, 5023, 7649, 12689, 13751, 18089, 19249, 19889, 22193, 22639, 23057, 23311, 23561, 26839, 27919, 28027, 28751, 30449, 30941, 31121, 32993, 33641, 33967, 36251, 38177, 38431, 39799, 39929

Here are the details:

  previous                prime   next

  2^5 * 59                1889    2 * 3^3 * 5 * 7
  2 * 3^4 * 19            3079    2^3 * 5 * 7 * 11
  2 * 3^3 * 5 * 17        4591    2^4 * 7 * 41
  2 * 3^4 * 31            5023    2^5 * 157
  2^5 * 239               7649    2 * 3^2 * 5^2 * 17
  2^4 * 13 * 61           12689   2 * 3^3 * 5 * 47
  2 * 5^4 * 11            13751   2^3 * 3^2 * 191
  2^3 * 7 * 17 * 19       18089   2 * 3^3 * 5 * 67
  2^4 * 3 * 401           19249   2 * 5^3 * 7 * 11
  2^4 * 11 * 113          19889   2 * 3^2 * 5 * 13 * 17
  2^4 * 19 * 73           22193   2 * 3^4 * 137
  2 * 3 * 7^3 * 11        22639   2^4 * 5 * 283
  2^4 * 11 * 131          23057   2 * 3^3 * 7 * 61
  2 * 3^2 * 5 * 7 * 37    23311   2^4 * 31 * 47
  2^3 * 5 * 19 * 31       23561   2 * 3^2 * 7 * 11 * 17
  2 * 3^3 * 7 * 71        26839   2^3 * 5 * 11 * 61
  2 * 3^3 * 11 * 47       27919   2^4 * 5 * 349
  2 * 3^4 * 173           28027   2^2 * 7^2 * 11 * 13
  2 * 5^4 * 23            28751   2^4 * 3 * 599
  2^4 * 11 * 173          30449   2 * 3 * 5^2 * 7 * 29
  2^2 * 5 * 7 * 13 * 17   30941   2 * 3^4 * 191
  2^4 * 5 * 389           31121   2 * 3^2 * 7 * 13 * 19
  2^5 * 1031              32993   2 * 3^3 * 13 * 47
  2^3 * 5 * 29^2          33641   2 * 3^3 * 7 * 89
  2 * 3^3 * 17 * 37       33967   2^4 * 11 * 193
  2 * 5^4 * 29            36251   2^2 * 3^2 * 19 * 53
  2^5 * 1193              38177   2 * 3^3 * 7 * 101
  2 * 3^2 * 5 * 7 * 61    38431   2^5 * 1201
  2 * 3^3 * 11 * 67       39799   2^3 * 5^2 * 199
  2^3 * 7 * 23 * 31       39929   2 * 3 * 5 * 11^3

The algorithm linked to above is easily modified to find 7-P-7 primes that are surrounded by two composite numbers with exactly seven prime factors with multiplicity. There are only four in the range up to 40000 are these are 10529, 15391, 32561 and 35153. The details are (permalink):

  previous            prime   next

  2^5 * 7 * 47        10529   2 * 3^4 * 5 * 13
  2 * 3^4 * 5 * 19    15391   2^5 * 13 * 37
  2^4 * 5 * 11 * 37   32561   2 * 3^5 * 67
  2^4 * 13^3          35153   2 * 3^4 * 7 * 31

There are no 8-P-8 primes in the range up to 40000 but if we extend the range to 100000 we find one (permalink):

  previous         prime   next

  2^6 * 11 * 107   75329   2 * 3^5 * 5 * 31

These number properties of certain primes are not base-dependent. Obviously as the numbers get bigger there will be instances of 9-P-9 primes and beyond.